Answer:
The coefficient of friction in the hall is 0.038
Explanation:
Given;
mass of the Parker, m = 73.2 kg
applied force on the parker, F = 123 N
frictional force, Fs = 27.4 N
the coefficient of friction in the hall = ?
frictional force is given by;
Fs = μN
Where;
μ is the coefficient of friction
N is normal reaction = mg
Fs = μmg
μ = Fs / mg
μ = (27.4) / (73.2 x 9.8)
μ = 0.038
Therefore, the coefficient of friction in the hall is 0.038
Answer:
v= 4 m/s
Explanation:
Momenutm is, by definition, the product of mass and velocity.

Let's replace what we know and solve for whatever's left

Explanation:
K.E =1/2 mv^2
=1/2(156789)(45.6)^2
=78,394.5 × 2,079.36
=163,010,387.52 kg m/s
This should be your answer.
Answer:
The component of F along AB is equal to Fcos45
F = 520N
Component along AB = 520cos45
= 367.7N
This is done by rotating the diamonds such that AB is now taken as the x-axis. Then the force F is resolved along AB.
Explanation:
<span>If the Earth rotated more slowly about its axis, your apparent weight would
A) increase.
B) decrease.
C) stay the same.
D) be zero.
</span>A) increase.